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Bases and Dimension

A basis β for a vector space V is a linearly independent subset of V
that generates or span V. If β is a basis for V, we also say that elements of β
form a basis for V. This means that every vector from V is a finite linear combination of elements
from the basis.

Recall that a set of vectors β is said to generate or span a vector space V if every element from
V can be represented as a linear combination of vectors from β.

Example: The span of the empty set \( \varnothing \) consists of a unique element 0.
Therefore, \( \varnothing \) is linearly independent and it is a basis for the trivial vector space
consisting of the unique element---zero. Its dimension is zero.

Example: Let us consider the set of all real \( m \times n \)
matrices, and let \( {\bf M}_{i,j} \) denote the matrix whose only nonzero entry is a 1 in
the i-th row and j-th column. Then the set \( {\bf M}_{i,j} \ : \ 1 \le i \le m , \ 1 \le j \le n \)
is a basis for the set of all such real matrices. Its dimension is mn.

Example: The set of monomials \( \left\{ 1, x, x^2 , \ldots , x^n \right\} \)
form a basis in the set of all polynomials of degree up to n. It has dimension n+1.
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Example: The infinite set of monomials \( \left\{ 1, x, x^2 , \ldots , x^n , \ldots \right\} \)
form a basis in the set of all polynomials.
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Theorem: Let V be a vector space and
\( \beta = \left\{ {\bf u}_1 , {\bf u}_2 , \ldots , {\bf u}_n \right\} \) be a subset of
V. Then β is a basis for V if and only if each vector v in V can be uniquely
decomposed into a linear combination of vectors in β, that is, can be uniquely expressed in the form

Let β be a basis for V. If \( {\bf v} \in V , \) then v belongs to the span of the basis β
because β is a basis for V, so it generates this vector space. Thus, v is a linear
combination of the elments of β. Suppose that

is uniquely expressed as a linear combination of elments from
β. It remains to prove that vectors from β are linearly independent because we know that the set of vectors
from β generates the vector space. Suppose that this statement is not true and one vector from β, say
u1 is expressed as a linear combination of other elements:

\[
{\bf u}_1 = a_2 {\bf u}_2 + \cdots + a_n {\bf u}_n .
\]

Then arbitrary vector v from V will have two distinct linear combination representations:

Theorem: Let S be a linearly independent subset of a vector space V,
and let v be an element of V that is not in S. Then
\( S \cup \{ {\bf v} \} \) is linearly dependent if and only if v
belongs to the span of the set S.

The next example demonstrates how Mathematica can determine the basis or set of linearly independent
vectors from the given set. Note that basis is not unique and even changing the order of vectors, a software can provide
you another set of linearly independent vectors.

In m1 you see 1 row and n columns together,so you can transpose it to see it as column
{{1, 1, 1, 3}, {0, 1, 0, 1}, {1, 1, 2, 4}}
One can use also the standard Mathematica command: IndependenceTest.
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A vector space is called finite-dimensional if it has a basis consisting of a finite number of
elements. The unique number of elements in each basis for V is called the dimension of
V and is denoted by dim(V). A vector space that is not finite-dimensional is called
infinite-dimensional.

Example: Over the field of complex numbers, the vector space \( \mathbb{C} \)
of all complex numbers has dimension 1 because its basis consists of one element \( \{ 1 \} . \)

Over the field of real numbers, the vector space \( \mathbb{C} \) of all complex numbers
has dimension 2 because its basis consists of two elements \( \{ 1, {\bf j} \} . \)